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- 84
- 120
- 60
- 53

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Solution

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Number of four digit numbers divisible by 2 or 5 should contain 0,2 or 5 in the end of the number.

→

Implies 4(3!) numbers.

Now eliminating the numbers with 0 at the starting, we get

4(3!)−3!

=3(3!)

=18. ...(i)

Implies 4(3!) numbers.

=24 ...(ii)

=24 ...(ii)

Implies 4(3!)

Eliminating the numbers with 0 at the starting, we get

4(3!)−3!

=3(3!)

=18 ...(ii)

Adding (i), (ii) and (iii), we get

18+24+18

=36+24

=60

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